# Suppose there existed a planet that went around the sun twice as fast as the earth. What would be...

## Question:

Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

## Kepler's third law:

Kepler's third law can be defined as the law which is applied for planetary revolution motion. As per this law, power two for the time required in the revolution is proportional to the third power of the length of the orbit.

## Answer and Explanation:

Let the time required by the earth to complete the rotation around sun is, {eq}T_e {/eq}.

and time required by the planet to complete the rotation around sun is, {eq}T_p {/eq}.

**Given data**

{eq}T_e =2 T_p {/eq}

The expression for Kepler?s third law is,

{eq}T^2 \propto r^3 {/eq}

Here T is time to complete one revolution.

r is total length of revolution.

Here for given case,

{eq}\left ( \frac{R_p}{R_e} \right )^3=\left ( \frac{T_p}{T_e} \right )^2 {/eq}

Substitute the given values,

{eq}\begin{align*}\left ( \frac{R_p}{R_e} \right )^3 &=\left ( \frac{T_p}{T_e} \right )^2\\\left ( \frac{R_p}{R_e} \right )^3 &=\left ( \frac{T_p}{2T_p} \right )^2\\\left ( \frac{R_p}{R_e} \right )&=(0.5)^{1.5}\\&=0.63 \end{align*} {/eq}

Thus the orbit of planet is 0.63 times of orbit of earth.

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